Question: Is ${818024}$ divisible by $3$ ?
A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {818024}= &&{8}\cdot100000+ \\&&{1}\cdot10000+ \\&&{8}\cdot1000+ \\&&{0}\cdot100+ \\&&{2}\cdot10+ \\&&{4}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {818024}= &&{8}(99999+1)+ \\&&{1}(9999+1)+ \\&&{8}(999+1)+ \\&&{0}(99+1)+ \\&&{2}(9+1)+ \\&&{4} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {818024}= &&\gray{8\cdot99999}+ \\&&\gray{1\cdot9999}+ \\&&\gray{8\cdot999}+ \\&&\gray{0\cdot99}+ \\&&\gray{2\cdot9}+ \\&& {8}+{1}+{8}+{0}+{2}+{4} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${818024}$ is divisible by $3$ if ${ 8}+{1}+{8}+{0}+{2}+{4}$ is divisible by $3$ Add the digits of ${818024}$ $ {8}+{1}+{8}+{0}+{2}+{4} = {23} $ If ${23}$ is divisible by $3$ , then ${818024}$ must also be divisible by $3$ ${23}$ is not divisible by $3$, therefore ${818024}$ must not be divisible by $3$.